Interpreting a Projectile Data Set with Air Resistance
Students fire a ball horizontally from a 1.20 m bench and record horizontal range for five launch speeds. Their data: at 4.0 m/s range 1.98 m; 6.0 m/s range 2.95 m; 8.0 m/s range 3.86 m; 10.0 m/s range 4.70 m; 12.0 m/s range 5.45 m. The theoretical no-drag model predicts range R = v√(2h/g). Students are asked to decide whether the data support the no-drag model and to identify the dominant source of discrepancy. They must also consider that the ball is a light plastic sphere of mass 2.6 g and diameter 4.0 cm.
1. Calculate the theoretical range at each speed and compare with the data to judge whether the no-drag model is supported.
Read the worked answer
Using R = v√(2h/g) with h = 1.20 m, √(2×1.20/9.8) = √0.245 = 0.495 s. Predicted ranges: 4.0→1.98 m; 6.0→2.97 m; 8.0→3.96 m; 10.0→4.95 m; 12.0→5.94 m. The measured values match at low speed (1.98 vs 1.98) but fall increasingly short: 2.95 vs 2.97, 3.86 vs 3.96, 4.70 vs 4.95, 5.45 vs 5.94. The discrepancy grows with speed, which is the signature of a velocity-dependent drag force rather than a constant offset. The no-drag model is only adequate at the lowest speed.
2. The ball is light and large. Explain why this matters and predict the effect of using a denser ball of the same size.
Read the worked answer
Drag force scales with cross-sectional area and v², while inertia scales with mass. A 2.6 g ball of 4.0 cm diameter has a large area-to-mass ratio, so drag decelerates it strongly, especially at high speed. A denser ball of the same size has the same drag but more mass, so the deceleration a = F_drag/m is smaller and the range is closer to the ideal. The data's growing shortfall is consistent with this. The dominant source of discrepancy is air resistance, not measurement error, because the shortfall is systematic and speed-dependent.
3. A student claims the shortfall is due to the bench height being measured too low. Test this claim quantitatively.
Read the worked answer
If the true height were larger, the predicted range would be larger, worsening the mismatch (predictions already exceed data at high speed). If the height were smaller, predictions would shrink. To match 5.45 m at 12.0 m/s, we need √(2h/g) = 5.45/12.0 = 0.454, so h = g(0.454)²/2 = 1.01 m. That is a 0.19 m error, far larger than plausible. Also, a height error would scale all predictions by the same factor, but the ratio measured/predicted falls from 1.00 to 0.92 as speed rises. A constant scaling cannot explain a speed-dependent trend, so the height claim fails.